Clearly the scheme given is not a hash function, in the sense that it is very easy to invert the function to obtain the original message. Hence, with any arbitrary hash, we can easily construct a message with that hash by inverting the operation. Python Implementation: from pwn import * import json BLOCK_SIZE = 32 # Nothing up my sleeve numbers (ref: Dual_EC_DRBG P-256 coordinates) W = [0x6b17d1f2, 0xe12c4247, 0xf8bce6e5, 0x63a440f2, 0x77037d81, 0x2deb33a0, 0xf4a13945, 0xd898c296] X = [0x4fe342e2, 0xfe1a7f9b, 0x8ee7eb4a, 0x7c0f9e16, 0x2bce3357, 0x6b315ece, 0xcbb64068, 0x37bf51f5] Y = [0xc97445f4, 0x5cdef9f0, 0xd3e05e1e, 0x585fc297, 0x235b82b5, 0xbe8ff3ef, 0xca67c598, 0x52018192] Z = [0xb28ef557, 0xba31dfcb, 0xdd21ac46, 0xe2a91e3c, 0x304f44cb, 0x87058ada, 0x2cb81515, 0x1e610046] # Lets work with bytes instead!

We will exploit the fact that numpy’s array class is int64. So, what we’re gonna do is basically generate random passwords, with digits, lowercase and uppercase letters and hope that one of these passwords have both sum and product (overflowed) prime. For this, we need password with three restrictions: It is large enough for the overflow to occur All the digits are odd (if we have any even digit, the product will be even, and therefore not prime) The length must be odd, so the sum will also be odd, as prime numbers should be odd numbers only I guess the last criteria is luck.

The vulnerability of the code (and yes, as somewhat hinted at by the challenge description), is these two lines: nonce = sha1(long_to_bytes(privkey.secret_multiplier) + hsh).digest() sig = privkey.sign(bytes_to_long(hsh), bytes_to_long(nonce)) sha1 produces a digest of only 160 bits (20 bytes). This is a big problem as it is required that the nonce is a number randomly generated in the range between 1 and the order of the elliptic curve. In the above code, the hash generated by sha1 is only 160 bits long.

There is a bug in the implementation, specifically def recover_x(self, xbit, y): xsqr = (y**2 - 1)*inverse(1 + self.d*y**2, self.p) % self.p x = pow(xsqr, (self.p + 1)//4, self.p) if x**2 == xsqr : ... return 0 the function recover_x will always return 0, as the check is not done on modulo $p$. The challenge is now straightforward as the base point has a x-coordinate of 0. Indeed, and referencing from this paper, the scalar multiplication on Edwards curve of a point $0, y$ is:

a and b is not given, but since we are given the two points, generator G and the public key point P, we can construct two equations in the form of $y ^ 2 = x ^ 3 + ax + b$. Solving the two linear congruent equation should give us $a, b$. Trying to do E = EllipticCurve(GF(p), [a, b]) lead to an error by Sage that the curve defined is not a elliptic curve, rather a singular curve.